# Maximal ideals of the rings of Baire-One Functions

A real function $$f:X\rightarrow \mathbb{R}$$ is called Baire-one function, if there is a sequence $$(f_n)_{n=1} ^\infty$$ of continuous functions $$f_n:X\rightarrow \mathbb{R}$$ on $$X$$ so that for all $$x\in X$$ $$\lim_{n\rightarrow \infty}f_n(x)=f(x).$$

When $$X$$ is a Banach space, we have the following theorem referred to as Baire factorization theorem.

Theorem: The real function $$f:X\rightarrow \mathbb{R}$$ is in the class of Baire-one if and only if for all closed subset $$K\subset X$$, the restricted function $$f|_K$$ has a point of continuity with respect to $$K$$.

Definition: We denote the set of all Baire-one real functions on the space $$X$$ by $$Ba_1(X)$$.

As you could easily see, $$Ba_1(X)$$ forms a ring with pointwise addition and multiplication. For simplicity let me consider $$X=[0 , 1]$$.

Suppose $$C[0 , 1]$$ denotes the ring of all continuous real valued functions on the interval $$[0 , 1]$$. By the theorem of Gelfond and Kolmogrov we know that the set of all maximal ideals of the ring $$C[0 , 1]$$ is of the form $$\{M_x: x\in X\}$$, where $$M_x=\{f\in C[0, 1]: f(x)=0\}$$.

Compared with the ring $$C[0 , 1]$$ we could easily find that the sets of the form $$M_x^1=\{f\in Ba_1[0 , 1]: f(x)=0\}$$ are maximal ideals of the ring $$Ba_1[0 , 1]$$. From this property some questions came in my mind as follows:

Question 1: Does there exist a maximal ideal in $$Ba_1[0 , 1]$$ other than maximal ideals of the form $$M_x^1$$ for $$x\in X$$?

Question 2: Is the ring $$Ba_1[0 , 1]$$ a $$\mathbf{PM}$$-ring? $$($$i.e. a ring in which each prime ideal is contained in a unique maximal ideal.$$)$$

• It looks to me like the collection of functions with finite support (that is, those which are zero on a cofinite set) forms an ideal not contained in any of the maximal ideals you've listed. If I understand your question correctly, that means the answer to question 1 is "yes." – Clinton Conley Aug 6 '12 at 23:00
• Yes Dear Clinton. Very Good. Also we could consider the ideal generated by all $\chi_{(a,1]}$ for $a \in (0,1]$, where $\chi_{(a,1]}$ is the characteristic function of the interval $(a , 1]$. then we could find a maximal ideal which contains all of these functions.I think this over ring of $C[0,1]$ has a complex behavior, because as you consider, it is full of idempotents. – Ali Reza Aug 6 '12 at 23:20
• Isn't $Ba_1(X)$ a commutative C*-algebra (in which case it is certainly a PM-ring)? – Douglas Somerset Mar 22 '13 at 22:43

## 1 Answer

The question may be answered in the comments. I thus collect the relevant comments here as an answer.

C. Conley: The collection of functions with finite support (that is, those which are zero on a cofinite set) forms an ideal not contained in any of the maximal ideals listed in the question.

D. Somerset: Perhaps $Ba_1(X)$ is a commutative C*-algebra, and thus a PM-ring.

Update: Y. Choi comments below that Somerset's approach does not work, since one is considering all the Baire-1 functions, not just the bounded ones.

• Douglas Somerset's comment (which also occurs in a deleted answer by someone else which is not visible unless one has sufficiently high reputation) is incorrect, because one is considering all the Baire-1 functions, not just the bounded ones. – Yemon Choi Aug 17 '15 at 3:05